Let's break down problem step by: ### **:** - of rod,L = \text{}\) - Initial diameter, \(D_1 = 6 \text{ cm}\) - Final diameter, \(D_2 = 5.4 \text{ cm}\) - Spindle rotates at \(N = 800 \text{ rpm}\) - Axial speed of tool, \(f = .30 \text{ cm/min}\) (assumed from "tool is traveling at an axial speed of 30 cm/min" but looks like a typo; should be 30 cm/min) ### **Material Removal Rate (MRR) formula for turning:** \[ \text{MRR} = \pi \times (D_1^2 - D_2^2) / 4 \times f \] Where: - \(D_1, D_2\) in cm - \(f\) is the feed speed in cm/min #### **Calculate:** \[ D_1^2 - D_2^2 = 6^2 - 5.4^2 = 36 - 29.16 = 6.84 \] \[ \text{MRR} = \frac{\pi}{4} \times 6.84 \times 30 \] \[ \frac{\pi}{4} \approx .7854 \] \[ \text{MRR} = .7854 \times 6.84 \times 30 \] \[ = .7854 \times 205.2 \] \[ = 161.15 \text{ cm}^3/\text{min} \] ### **Correct Option:** - **161 cm³/min** So, the correct answer is: > **161 cm³/min**